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Using the marginals we compute means E(X) = 0 and E(Y) = 2 Next we show that Xand Y are not independent To do this all we have to do is nd one place where the product rule fails, ie where p(x i;y j) 6= p(x i)p(x j) P(X= 2;Y = 0) = 0 but P(X= 2) P(Y = 0) = 1=25 Since these are not equal Xand Y are not independent Finally we computeIn fact, provided the condition above holds (ie, there exists a feasible x with ¯cTx ≥ α) we can solve the problem (1) via convex optimization We make the change of variables y = x ¯cTx−α, s = 1 ¯cTx−α, so x = y/s This yields the problem minimize q yTRy subject to Fy gs Ay = bs ¯cTy −αs = 1Yex xsiny g(y) = ex xcosy It follows that g0(y) = 0 and hence g(y) = K Therefore f(x,y) = yex xsiny K 6 Evaluate Z C Fdr along the given curve C (a) F(x,y) = y2 1x2 i 2yarctanx j, C r(t) = (t2) i(2t) j, 0 ≤ t ≤ 1 (b) F(x,y,z) = (y2 cosz)i(2xycosz)j−(xy2 sinz)k, C r(t) = (t2) i(sint) jt k, 0 ≤ t ≤ π Solution



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KCET 19 Maths question paper with solutions and answer key are available on this page Students will benefit from the step by step solutions prepared by our subject matter experts and gain a better understanding of the KCET level Maths problems and how to solve themA = (Ax cos <j>X W jˇ X F j r j Now, as usual, we let the pieces get in nitesimally small, so the sum becomes an integral and the approximation becomes exact We get total work = Z C Fdr The subscript Cindicates that it is the curve that has been split into pieces That is, the total work is computed as a line integral of the force over the curve C!



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